The volume V of a given mass of monoatomic gas changes with temperature T according to the relation `V=KT^(2//3)`. The workdone when temperature changes by 90 K will be xR. The value of x is [R = universal gas constant]

A gas expands with temperature according to the relation V=KT^(2/3) .Work done when the temperature changes by 60K is.

For one mole of an ideal monoatomic gas, volume and temperature are related as V = KT^(2/3) . If change in temperature of gas is 90K, then work done is given by W = xR. Find value of x

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT= K, where K is a constant. In this process the temperataure of the gas is increased by DeltaT . The amount of heat absorbed by gas is (R is gas constant).

One mole of an ideal monoatomic gas at temperature T_0 expands slowly according to the law p/V = constant. If the final temperature is 2T_0 , heat supplied to the gas is

One mole of an ideal monoatomic gas at temperature T_(0) expands slowely according to the law P/V=constant. If the final temperature is 2T_(0) heat supplied to the gas is -

The volume of a given mass of a gas in a vessel is reduced to half at constant temperature what is the change in pressure?

One mole of a gas expands with temperature T such thaht its volume, V= KT^(2) , where K is a constant. If the temperature of the gas changes by 60^(@)C then the work done by the gas is 120R

Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300K, it expands to a volume 4V adiabatically, then the final temperature of gas is